`sqrt(3)" cosec "20^(@)-sec20^(2)=?`

Options

A. 1
B. 2
C. 4
D. None of these

Answer

To solve the expression \( \sqrt{3} \csc 20^\circ - \sec 20^\circ \), we will follow these steps:

### Step 1: Rewrite the trigonometric functions
We know that:
- \( \csc 20^\circ = \frac{1}{\sin 20^\circ} \)
- \( \sec 20^\circ = \frac{1}{\cos 20^\circ} \)

So we can rewrite the expression as:
\[ \sqrt{3} \csc 20^\circ - \sec 20^\circ = \sqrt{3} \cdot \frac{1}{\sin 20^\circ} - \frac{1}{\cos 20^\circ} \]

### Step 2: Find a common denominator
The common denominator for the two fractions is \( \sin 20^\circ \cos 20^\circ \). Thus, we can write:
\[ \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ} = \frac{\sqrt{3} \cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ} \]

### Step 3: Simplify the numerator
Now we focus on simplifying the numerator \( \sqrt{3} \cos 20^\circ - \sin 20^\circ \). We can recognize that this resembles the sine of a sum or difference. We can express it as:
\[ \sqrt{3} \cos 20^\circ - \sin 20^\circ = 2 \left( \frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ \right) \]
This can be rewritten using the sine angle subtraction formula:
\[ = 2 \sin\left(60^\circ - 20^\circ\right) = 2 \sin 40^\circ \]

### Step 4: Substitute back into the expression
Now substituting back into our expression gives:
\[ \frac{2 \sin 40^\circ}{\sin 20^\circ \cos 20^\circ} \]

### Step 5: Use the double angle identity
We know that:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
Thus, we can express \( \sin 20^\circ \cos 20^\circ \) as:
\[ \sin 20^\circ \cos 20^\circ = \frac{1}{2} \sin 40^\circ \]

### Step 6: Final simplification
Now substituting this back gives:
\[ \frac{2 \sin 40^\circ}{\frac{1}{2} \sin 40^\circ} = \frac{2 \sin 40^\circ \cdot 2}{\sin 40^\circ} = 4 \]

### Final Answer
Thus, the value of \( \sqrt{3} \csc 20^\circ - \sec 20^\circ \) is:
\[ \boxed{4} \]